Problem: Let $y=\cos(x)x^3$. $\dfrac{dy}{dx}=$
Solution: $\cos(x)x^3$ is the product of two, more basic, expressions: $\cos(x)$ and $x^3$. Therefore, $\dfrac{dy}{dx}$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left[\cos(x)x^3\right] \\\\ &=\dfrac{d}{dx}\left[\cos(x)\right]x^3+\cos(x)\dfrac{d}{dx}[x^3]&&\gray{\text{The product rule}} \\\\ &=-\sin(x)x^3+\cos(x)3x^2&&\gray{\text{Differentiate }\cos(x)\text{ and }x^3} \\\\ &=3x^2\cos(x)-x^3\sin(x)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=3x^2\cos(x)-x^3\sin(x)$ or any other equivalent form.